Motion in Two Dimensions Chapter 6. Projectile Motion Recognize that the vertical and horizontal...
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Transcript of Motion in Two Dimensions Chapter 6. Projectile Motion Recognize that the vertical and horizontal...
Motion in Two DimensionsChapter
6
Projectile Motion
Recognize that the vertical and horizontal motions of a projectile are independent.
Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, and then determine the range using the horizontal motion.
Explain how the trajectory of a projectile depends upon the frame of reference from which it is observed.
In this section you will:
Section
6.1
Projectile Motion
If you observed the movement of a golf ball being hit from a tee, a frog hopping, or a free throw being shot with a basketball, you would notice that all of these objects move through the air along similar paths, as do baseballs, arrows, and bullets.
Each path is a curve that moves upward for a distance, and then, after a time, turns and moves downward for some distance.
You may be familiar with this curve, called a parabola
Projectile Motion
Section
6.1
Projectile Motion
An object shot through the air is called a projectile.
Projectile Motion
Section
6.1
A projectile can be a football, a bullet, or a drop of water.
The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory.
Independence of Motion in Two Dimensions
Section
6.1 Projectile Motion
Projectile Motion
When a projectile is launched at an angle, the initial velocity has a vertical component as well as a horizontal component.
If the object is launched upward, like a ball tossed straight up in the air, it rises with slowing speed, reaches the top of its path, and descends with increasing speed.
Projectiles Launched at an Angle
Section
6.1
Projectile Motion
At each point in the vertical direction, the velocity of the object as it is moving upward has the same magnitude as when it is moving downward.
The only difference is that the directions of the two velocities are opposite.
Projectiles Launched at an Angle
Section
6.1
Projectile Motion
The adjoining figure defines two quantities associated with a trajectory.
One is the maximum height, which is the height of the projectile when the vertical velocity is zero.
Projectiles Launched at an Angle
Section
6.1
Projectile Motion
The other quantity depicted is the range, R, which is the horizontal distance that the projectile travels.
Time is how much time the projectile is in the air.
Flight time often is called hang time.
Projectiles Launched at an Angle
Section
6.1
Choosing a coordinate system, such as the one shown below, is similar to laying a grid.
You have to choose where to put the center of the grid (the origin) and establish the directions in which the axes point.
You then have X’s and Y’s for components.
Vectors
Components of Vectors
Section
5.1
Vectors
When the motion is on the surface of Earth, it is often convenient to have the x-axis point east and the y-axis point north.
When the motion involves an object moving through the air, the positive x-axis is often chosen to be horizontal to the right and the positive y-axis vertical (upward).
If the motion is on a hill, it’s convenient to place the positive x-axis in the direction of the motion (along the ramp) and the y-axis perpendicular to the x-axis (normal)
Components of Vectors
Section
5.1
Vectors
Component Vectors
Section
5.1
Two or more vectors (A, B, C, etc.) may be added by first resolving each vector into its x- and y-components.
The x-components are added to form the x-component of the resultant: Rx = Ax + Bx + Cx.
Similarly, the y-components are added to form the y-component of the resultant:
Ry = Ay + By + Cy.
Vectors
Algebraic Addition of Vectors
Section
5.1
Vectors
Because Rx and Ry are at a right angle (90°), the magnitude of the resultant vector can be calculated using the Pythagorean theorem, R2 = Rx
2 + Ry2.
Algebraic Addition of Vectors
Section
5.1
Vectors
To find the angle or direction of the resultant, recall that the tangent of the angle that the vector makes with the x-axis is give by the following.
Angle of the resultant vector
Algebraic Addition of Vectors
Section
5.1
-1 y
x
= tanR
R
The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
Vectors
You can find the angle by using the tan−1 key on your calculator.
Note that when tan θ > 0, most calculators give the angle between 0° and 90°, and when tan θ < 0, the angle is reported to be between 0° and −90°.
You will use these techniques to resolve vectors into their components throughout your study of physics.
Resolving vectors into components allows you to analyze complex systems of vectors without using graphical methods.
Algebraic Addition of Vectors
Section
5.1
Projectile Motion
Angle - SOH, CAH, TOA
Magnitude – Hypotenuse
C2 = A2 + B2
Or R2 = Rx2 + Ry
2
Magnitude – Component
Rx = Rcosθ
Ry = Rsinθ
Vector Components – Trig Review (or new material)Your book – Chapter 5.1
Polar CS vs. Cartesian CS
Section
5.1
-1 y
x
= tanR
R
Vectors
Vector Direction – Cardinal Directions
Section
5.1
N
S
EW
N of E
E of N
S of W
W of S
N of W
W of N
S of E
E of S
Section Check
Jeff moved 3 m due north, and then 4 m due west to his friends house. What is the displacement of Jeff?
Question 1
Section
5.1
A. 3 + 4 m
B. 4 – 3 m
C. 32 + 42 m
D. 5 m
Section Check
Answer: D
Answer 1
Section
5.1
Reason: When two vectors are at right angles to each other as in this case, we can use the Pythagorean theorem of vector addition to find the magnitude of resultant, R.
Section Check
Answer: D
Answer 1
Section
5.1
Reason: Pythagorean theorem of vector addition states
If vector A is at right angle to vector B then the sum of squares of magnitudes is equal to square of magnitude of resultant vector.
That is, R2 = A2 + B2
R2 = (3 m)2 + (4 m)2 = 5 m
Calculate the resultant of three vectors A, B, C as shown in the figure.
(Ax = Bx = Cx = Ay = Cy = 1 units and By = 2 units)
2 24 + 3 units
Question 2
Section
5.1 Section Check
A. 3 + 4 units
B. 32 + 42 units
C. 42 – 32 units
D.
Section Check
Answer: D
Answer 2
Section
5.1
Reason: Add the x-components to form, Rx = Ax + Bx + Cx.
Add the y-components to form, Ry = Ay + By + Cy.
Section Check
Answer: D
Answer 2
Section
5.1
Reason: Since Rx and Ry are perpendicular to each other we can apply Pythagoras theorem of vector addition:
2 2 = 4 + 3 unitsR
R2 = Rx2 + Ry
2
Section Check
If a vector B is resolved into two components Bx and By and if is
angle that vector B makes with the positive direction of x-axis, Using which of the following formula can you calculate the components of vector B?
Question 3
Section
5.1
A. Bx = B sin θ, By = B cos θ
B. Bx = B cos θ, By = B sin θ
C.
D.
x y
cos sin = , = B BB B
x y
sin cos = , = B BB B
Section Check
Answer: B
Answer 3
Section
5.1
Reason: The components of vector ‘B’ are calculated using the equation stated below.
opposite sidehypotenuse
sin =
y sin = B
B
y = sin B B
Section Check
Answer: B
Answer 3
Section
5.1
Reason: Adjacent sidehypotenuse
Also cos =
x cos = BB
B Bx = cos
Projectile Motion
The Flight of a Ball
A ball is launched at 4.5 m/s at 66° above the horizontal.What are the maximum height and flight time of the ball?
Section
6.1
The Flight of a Ball
Establish a coordinate system with the initial position of the ball at the origin.
Projectile MotionSection
6.1
The Flight of a Ball
Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight.
Projectile MotionSection
6.1
The Flight of a Ball
Draw a motion diagram showing v, a, and Fnet.
Projectile MotionSection
6.1
Known:
yi = 0.0 m
θi = 66°
vi = 4.5 m/s
ay = −g
Unknown:
ymax = ?
t = ?
The Flight of a Ball
Find the y-component of vi.
Projectile MotionSection
6.1
The Flight of a Ball
Substitute vi = 4.5 m/s, θi = 66°
Projectile MotionSection
6.1
The Flight of a Ball
Find an expression for time.
Projectile MotionSection
6.1
The Flight of a Ball
Substitute ay = −g
Projectile MotionSection
6.1
The Flight of a Ball
Solve for t.
Section
6.1 Projectile Motion
The Flight of a Ball
Solve for the maximum height.
Section
6.1 Projectile Motion
The Flight of a Ball
Section
6.1
-Substitute , = -yi y
v vt = a g
g
yi y yi ymax i yi
– –1(– )
2
2
v v v v
y = y +v + gg g
Projectile Motion
The Flight of a Ball
Substitute yi = 0.0 m, vyi = 4.1 m/s, vy = 0.0 m/s at ymax, g = 9.80 m/s2
Projectile MotionSection
6.1
The Flight of a Ball
Solve for the time to return to the launching height.
Proof – Quadratic Formula
Projectile MotionSection
6.1
The Flight of a Ball
Substitute yf = 0.0 m, yi = 0.0 m, a = −g
Projectile MotionSection
6.1
The Flight of a Ball
Use the quadratic formula to solve for t.
Projectile MotionSection
6.1
2
yi yi
1– ± –4 –2= (0.0 m)
12 –2
v v gt
g
The Flight of a Ball
0 is the time the ball left the launch, so use this solution.
Projectile MotionSection
6.1
The Flight of a Ball
Substitute vyi = 4.1 m/s, g = 9.80 m/s2
Projectile MotionSection
6.1
Projectile Motion
The path of the projectile, or its trajectory, depends upon who is viewing it.
Suppose you toss a ball up and catch it while riding in a bus. To you, the ball would seem to go straight up and straight down.
But an observer on the sidewalk would see the ball leave your hand, rise up, and return to your hand, but because the bus would be moving, your hand also would be moving. The bus, your hand, and the ball would all have the same horizontal velocity.
Trajectories Depend upon the Viewer
Section
6.1
Section Check
A boy standing on a balcony drops one ball and throws another with an initial horizontal velocity of 3 m/s. Which of the following statements about the horizontal and vertical motions of the balls is correct? (Neglect air resistance.)
Question 1
Section
6.1
A. The balls fall with a constant vertical velocity and a constant horizontal acceleration.
B. The balls fall with a constant vertical velocity as well as a constant horizontal velocity.
C. The balls fall with a constant vertical acceleration and a constant horizontal velocity.
D. The balls fall with a constant vertical acceleration and an increasing horizontal velocity.
Section Check
Answer: C
Answer 1
Section
6.1
Reason: The vertical and horizontal motions of a projectile are independent. The only force acting on the two balls is force due to gravity. Because it acts in the vertical direction, the balls accelerate in the vertical direction. The horizontal velocity remains constant throughout the flight of the balls.
Section Check
Which of the following conditions is met when a projectile reaches its maximum height?
Question 2
Section
6.1
A. Vertical component of the velocity is zero.
B. Vertical component of the velocity is maximum.
C. Horizontal component of the velocity is maximum.
D. Acceleration in the vertical direction is zero.
Section Check
Answer: A
Answer 2
Section
6.1
Reason: The maximum height is the height at which the object stops its upward motion and starts falling down, i.e. when the vertical component of the velocity becomes zero.
Section Check
Suppose you toss a ball up and catch it while riding in a bus. Why does the ball fall in your hands rather than falling at the place where you tossed it?
Question 3
Section
6.1
Section Check
No air resistance – nothing to slow it down.
Trajectory depends on the frame of reference.
Also inertia (Newton’s Laws)
Answer 3
Section
6.1
Circular Motion
This section will not be covered at this time.
Section
6.2
Relative Velocity
Analyze situations in which the coordinate system is moving.
Solve relative-velocity problems.
In this section you will:
Section
6.3
Relative Velocity
You are in a school bus that is traveling at a velocity of 8 m/s. You walk with a velocity of 3 m/s.
If the bus is traveling at 8 m/s, this means that the velocity of the bus is 8 m/s, as measured by your friend in a coordinate system fixed to the road.
Relative Velocity
Section
6.3
Relative Velocity
When you are standing still, your velocity relative to the road is also 8 m/s, but your velocity relative to the bus is zero.
A vector representation of this problem is shown in the figure.
Relative Velocity
Section
6.3
Relative Velocity
When a coordinate system is moving, two velocities are added if both motions are in the same direction.
One is subtracted from the other if the motions are in opposite directions.
You will find that your velocity relative to the street is 11 m/s, the sum of 8 m/s and 3 m/s.
From the frame of reference to someone in the bus you are moving 3 m/s.
Relative Velocity
Section
6.3
Relative Velocity
Two velocities are in opposite directions, the resultant velocity is 5 m/s–the difference between 8 m/s and 3 m/s.
Simple addition or subtraction can be used to determine the relative velocity.
Relative Velocity
Section
6.3
Relative Velocity
Mathematically, relative velocity is represented as vy/b + vb/r = vy/r.
Y=you, b=bus, and r=road
The more general form of this equation is:
Relative Velocity
Section
6.3
The method for adding relative velocities also applies to motion in two dimensions.
Relative Velocity
Relative Velocity
Section
6.3
Airline pilots must take into account the plane’s speed relative to the air, and their direction of flight relative to the air.
They also must consider the velocity of the wind at the altitude they are flying relative to the ground.
Relative Velocity
Relative Velocity of a Marble
Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water?
Section
6.3
Relative Velocity of a Marble
Establish a coordinate system.
Relative VelocitySection
6.3
Relative Velocity of a Marble
Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat.
Relative VelocitySection
6.3
Relative Velocity of a Marble
Identify known and unknown variables.
Relative VelocitySection
6.3
Known:
vb/w = 4.0 m/s
vm/b = 0.75 m/s
Unknown:
vm/w = ?
Relative Velocity of a Marble
Because the two velocities are at right angles, use the Pythagorean theorem.
Relative VelocitySection
6.3
Relative Velocity of a Marble
Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s
Relative VelocitySection
6.3
Relative Velocity of a Marble
Find the angle of the marble’s motion.
Relative VelocitySection
6.3
Relative Velocity of a Marble
Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s
Relative VelocitySection
6.3
= 11° north of east
The marble is traveling 4.1 m/s at 11° north of east.
Relative Velocity
Are the units correct?
Dimensional analysis verifies that the velocity is in m/s.
Do the signs make sense?
The signs should all be positive.
Are the magnitudes realistic?
The resulting velocity is of the same order of magnitude as the velocities given in the problem.
Relative Velocity of a Marble
Section
6.3
Relative Velocity
You can add relative velocities even if they are at arbitrary angles by using the graphical methods.
Analyzing a two-dimensional relative- velocity situation is drawing the proper triangle to represent the velocities. Once you have this triangle, you simply apply your knowledge of vector addition.
If two velocities are perpendicular to each other, you can find the third by applying the Pythagorean theorem.
If the situation has no right angles, you will need to use one or both of the laws of sines and cosines. You will rarely have to do this in HS physics.
Relative Velocity
Section
6.3
Section Check
Steven is walking on the roof of a bus with a velocity of 2 m/s toward the rear end of the bus. The bus is moving with a velocity of 10 m/s. What is the velocity of Steven with respect to Anudja sitting inside the bus and Mark standing on the street?
Question 1
Section
6.3
A. Velocity of Steven with respect to Anudja is 2 m/s and with respect to Mark is 12 m/s.
B. Velocity of Steven with respect to Anudja is 2 m/s and with respect to Mark is 8 m/s.
C. Velocity of Steven with respect to Anudja is 10 m/s and with respect to Mark is 12 m/s.
D. Velocity of Steven with respect to Anudja is 10 m/s and with respect to Mark is 8 m/s.
Section Check
Answer: B
Answer 1
Section
6.3
Reason: The velocity of Steven with respect to Anudja is 2 m/s since Steven is moving with a velocity of 2 m/s with respect to the bus, and Anudja is at rest with respect to the bus.
The velocity of Steven with respect to Mark can be understood with the help of the following vector representation.
Section Check
Which of the following formulas is the general form of relative velocity of objects a, b, and c?
Question 2
Section
6.3
A. Va/b + Va/c = Vb/c
B. Va/b Vb/c = Va/c
C. Va/b + Vb/c = Va/c
D. Va/b Va/c = Vb/c
Section Check
Answer: C
Answer 2
Section
6.3
Reason: Relative velocity law is Va/b + Vb/c = Va/c.
The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.
Section Check
An airplane flies due south at 100 km/hr relative to the air. Wind is blowing at 20 km/hr to the west relative to the ground. What is the plane’s speed with respect to the ground?
Question 3
Section
6.3
A. (100 + 20) km/hr
B. (100 − 20) km/hr
C.
D.
Section Check
Answer: C
Answer 3
Section
6.3
Reason: Since the two velocities are at right angles, we can apply Pythagoras theorem of addition law. By using relative velocity law, we can write:
End of Chapter
6Chapte
r Motion in Two Dimensions